A remark on partial sums involving the Mobius function

Details A remark on partial sums involving the Mobius function A remark on partial sums involving the Mobius function

2009-08-29

arxiv.org/abs/0908.4323v5

Mathematics

Number Theory

7 pages, no figures. To appear, Bull. Aust. Math. Soc. Minor corrections

Scientific paper

Let $<\P > \subset \N$ be a multiplicative subsemigroup of the natural numbers $\N = \{1,2,3,...\}$ generated by an arbitrary set $\P$ of primes (finite or infinite). We given an elementary proof that the partial sums $\sum_{n \in < \P >: n \leq x} \frac{\mu(n)}{n}$ are bounded in magnitude by 1. With the aid of the prime number theorem, we also show that these sums converge to $\prod_{p \in \P} (1 - \frac{1}{p})$ (the case when $\P$ is all the primes is a well-known observation of Landau). Interestingly, this convergence holds even in the presence of non-trivial zeroes and poles of the associated zeta function $\zeta_\P(s) := \prod_{p \in \P} (1-\frac{1}{p^s})^{-1}$ on the line $\{\Re(s)=1\}$. As equivalent forms of the first inequality, we have $|\sum_{n \leq x: (n,P)=1} \frac{\mu(n)}{n}| \leq 1$, $|\sum_{n|N: n \leq x} \frac{\mu(n)}{n}| \leq 1$, and $|\sum_{n \leq x} \frac{\mu(mn)}{n}| \leq 1$ for all $m,x,N,P \geq 1$.

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